USING MEAN ABSOLUTE DEVIATION

About "Using mean absolute deviation"

The mean absolute deviation can be used to answer statistical questions in the real world. Many of these questions may have implications for the operation of various businesses.

Using Mean Absolute Deviation - Example

A chicken farmer wants her chickens to all have about the same weight. She is trying two types of feed to see which type produces the best results. All the chickens in Pen A are fed Premium Growth feed, and all the chickens in Pen B are fed Maximum Growth feed. The farmer records the weights of the chickens in each pen in the tables below. Which chicken feed produces less variability in weight ?

Solution :

Step 1 :

Find the mean weight of the chickens in each pen. Round your answers to the nearest tenth.

Pen A = (5.8+6.1+5.5+6.6+7.3+5.9+6.3+5.7+6.8+7.1) / 10

Pen A = 63.1 / 10

Pen A = 6.31

Pen A ≈ 6.3

Pen B = (7.7+7.4+5.4+7.8+6.1+5.2+7.5+7.9+6.3+5.6) / 10

Pen A = 66.9 / 10

Pen A = 6.69

Pen A ≈ 6.7

Step 2 :

Find the absolute deviations from the mean for each of the weights.

The absolute deviations from the mean for Pen A are the distance of each weight from 6.3 lb.

|5.8 - 6.3| = |-0.5| = 0.5

|6.1 - 6.3| = |-0.2| = 0.2

|5.5 - 6.3| = |-0.8| = 0.8

|6.6 - 6.3| = |0.3| = 0.3

|7.3 - 6.3| = |1.0| = 1.0

|5.9 - 6.3| = |-0.4| = 0.4

|6.3 - 6.3| = |0| = 0

|5.7 - 6.3| = |-0.6| = 0.6

|6.8 - 6.3| = |0.5| = 0.5

|7.1 - 6.3| = |0.8| = 0.8

The absolute deviations from the mean for Pen B are the distance of each weight from 6.7 lb.

|7.7 - 6.7| =
|1.0| = 1.0

|7.4 - 6.7|
= |0.7| = 0.7

|5.4 - 6.7|
= |-1.3| = 1.3

|7.8 - 6.7|
= |1.1| = 1.1

|6.1 - 6.7|
= |-0.6| = 0.6

|5.2 - 6.7|
= |-1.5| = 1.5

|7.5 - 6.7|
= |0.8| = 0.8

|7.9 - 6.7|
= |1.2| = 1.2

|6.3 - 6.7|
= |-0.4| = 0.4

|5.6 - 6.7|
= |-1.1| = 1.1

Step 3 :

Calculate the mean absolute deviation for the
chickens in each pen. Round your answers to the nearest tenth.